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| Derivation / Formula | Reasoning |
|---|---|
| \[\theta = 160\,(2\pi) = 320\pi \;\text{rad}\] | The total angular displacement is the number of revolutions multiplied by \(2\pi\) radians per revolution. |
| \[\alpha = \frac{2\theta}{t^{2}}\] | For uniformly accelerated rotation starting from rest, \(\theta = \tfrac{1}{2}\alpha t^{2}\); solving for \(\alpha\) gives this expression. |
| \[\alpha = \frac{2(320\pi)}{(15.0)^{2}} = 8.94\;\text{rad\,s}^{-2}\] | Substitute \(\theta = 320\pi\,\text{rad}\) and \(t = 15.0\,\text{s}\) to obtain the angular acceleration. |
| \[I = \frac{\tau}{\alpha}\] | Newton’s second law for rotation: torque \(\tau\) equals moment of inertia \(I\) times angular acceleration \(\alpha\). |
| \[I = \frac{10.8\,\text{N\,m}}{8.94\,\text{rad\,s}^{-2}} = 1.21\;\text{kg\,m}^{2}\] | Insert \(\tau = 10.8\,\text{N·m}\) and \(\alpha = 8.94\,\text{rad\,s}^{-2}\) to find the sphere’s moment of inertia. |
| \[I = \frac{2}{5} M R^{2}\] | Moment of inertia for a solid sphere of mass \(M\) and radius \(R\). |
| \[M = \frac{5I}{2R^{2}}\] | Solve the solid-sphere moment-of-inertia formula for mass \(M\). |
| \[M = \frac{5(1.21)}{2(0.36)^{2}} = 23.3\;\text{kg}\] | Substitute \(I = 1.21\,\text{kg\,m}^{2}\) and \(R = \tfrac{0.72}{2} = 0.36\,\text{m}\) to calculate the mass. |
| \[\boxed{M \approx 23.3\,\text{kg}}\] | The required mass of the solid sphere. |
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The diagram above shows a top view of a child of mass \(M\) on a circular platform of mass \(2M\) that is rotating counterclockwise. Assume the platform rotates without friction. Which of the following describes an action by the child that will increase the angular speed of the platform-child system and gives the correct reason why?
A solid sphere, solid cylinder, and a hollow pipe all have equal masses and radii. If the three of them are released simultaneously at the top of an inclined plane and do not slip, which one will reach the bottom first? [katex] I_{sphere} = \frac{2}{5}MR^2[/katex], [katex] I_{cylinder} = \frac{1}{2}MR^2[/katex], [katex] I_{pipe} = MR^2[/katex]
A solid metal bar is at rest on a horizontal frictionless surface. It is free to rotate about a vertical axis at the left end. The figures below show forces of different magnitudes that are exerted on the bar at different locations. In which case does the bar’s angular speed about the axis increase at the fastest rate?
An old record player could bring a disk up to its \(45\) RPM speed in less than a second. If the same size disk can also be brought up to a speed of \(75\) RPM in about the same amount of time on another player. Compare the torques exerted by each record player.
If a constant net torque is applied to an object it will (select all that applies):
In both cases, a massless rod is supported by a fulcrum, and a \(200 \, \text{kg}\) hanging mass is suspended from the left end of the rod by a cable. A downward force \(F\) keeps the rod in rest. The rod in Case A is \(50 \, \text{cm}\) long, and the rod in Case B is \(40 \, \text{cm}\) long (each rod is marked at \(10 \, \text{cm}\) intervals). The magnitude of each vertical force \(F\) exerted on the rod will be
In the figure above, the marble rolls down the track and around a loop-the-loop of radius \( R \). The marble has mass \( m \) and radius \( r \). What minimum height \( h_{min} \) must the track have for the marble to make it around the loop-the-loop without falling off? Express your answer in terms of the variables \( R \) and \( r \).
To increase the moment of inertia of a body about an axis, you must
A \( 4 \)-\( \text{kg} \) ball and a \( 1 \)-\( \text{kg} \) ball are positioned a distance \( L \) apart on a bar of negligible mass. How far from the \( 4 \)-\( \text{kg} \) mass should the fulcrum be placed to balance the bar?
A windmill blade with a rotational inertia of \( 6.0 \) \( \text{kg} \cdot \text{m}^2 \) has an initial angular velocity of \( 8 \) \( \text{rad/s} \) in the clockwise direction. It is then given an angular acceleration of \( 4 \) \( \text{rad/s}^2 \) in the clockwise direction for \( 10 \) seconds. What is the change in rotational kinetic energy of the blade over this time interval?
\(23.3\,\text{kg}\)
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| Kinematics | Forces |
|---|---|
| \(\Delta x = v_i t + \frac{1}{2} at^2\) | \(F = ma\) |
| \(v = v_i + at\) | \(F_g = \frac{G m_1 m_2}{r^2}\) |
| \(v^2 = v_i^2 + 2a \Delta x\) | \(f = \mu N\) |
| \(\Delta x = \frac{v_i + v}{2} t\) | \(F_s =-kx\) |
| \(v^2 = v_f^2 \,-\, 2a \Delta x\) |
| Circular Motion | Energy |
|---|---|
| \(F_c = \frac{mv^2}{r}\) | \(KE = \frac{1}{2} mv^2\) |
| \(a_c = \frac{v^2}{r}\) | \(PE = mgh\) |
| \(T = 2\pi \sqrt{\frac{r}{g}}\) | \(KE_i + PE_i = KE_f + PE_f\) |
| \(W = Fd \cos\theta\) |
| Momentum | Torque and Rotations |
|---|---|
| \(p = mv\) | \(\tau = r \cdot F \cdot \sin(\theta)\) |
| \(J = \Delta p\) | \(I = \sum mr^2\) |
| \(p_i = p_f\) | \(L = I \cdot \omega\) |
| Simple Harmonic Motion | Fluids |
|---|---|
| \(F = -kx\) | \(P = \frac{F}{A}\) |
| \(T = 2\pi \sqrt{\frac{l}{g}}\) | \(P_{\text{total}} = P_{\text{atm}} + \rho gh\) |
| \(T = 2\pi \sqrt{\frac{m}{k}}\) | \(Q = Av\) |
| \(x(t) = A \cos(\omega t + \phi)\) | \(F_b = \rho V g\) |
| \(a = -\omega^2 x\) | \(A_1v_1 = A_2v_2\) |
| Constant | Description |
|---|---|
| [katex]g[/katex] | Acceleration due to gravity, typically [katex]9.8 , \text{m/s}^2[/katex] on Earth’s surface |
| [katex]G[/katex] | Universal Gravitational Constant, [katex]6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2[/katex] |
| [katex]\mu_k[/katex] and [katex]\mu_s[/katex] | Coefficients of kinetic ([katex]\mu_k[/katex]) and static ([katex]\mu_s[/katex]) friction, dimensionless. Static friction ([katex]\mu_s[/katex]) is usually greater than kinetic friction ([katex]\mu_k[/katex]) as it resists the start of motion. |
| [katex]k[/katex] | Spring constant, in [katex]\text{N/m}[/katex] |
| [katex] M_E = 5.972 \times 10^{24} , \text{kg} [/katex] | Mass of the Earth |
| [katex] M_M = 7.348 \times 10^{22} , \text{kg} [/katex] | Mass of the Moon |
| [katex] M_M = 1.989 \times 10^{30} , \text{kg} [/katex] | Mass of the Sun |
| Variable | SI Unit |
|---|---|
| [katex]s[/katex] (Displacement) | [katex]\text{meters (m)}[/katex] |
| [katex]v[/katex] (Velocity) | [katex]\text{meters per second (m/s)}[/katex] |
| [katex]a[/katex] (Acceleration) | [katex]\text{meters per second squared (m/s}^2\text{)}[/katex] |
| [katex]t[/katex] (Time) | [katex]\text{seconds (s)}[/katex] |
| [katex]m[/katex] (Mass) | [katex]\text{kilograms (kg)}[/katex] |
| Variable | Derived SI Unit |
|---|---|
| [katex]F[/katex] (Force) | [katex]\text{newtons (N)}[/katex] |
| [katex]E[/katex], [katex]PE[/katex], [katex]KE[/katex] (Energy, Potential Energy, Kinetic Energy) | [katex]\text{joules (J)}[/katex] |
| [katex]P[/katex] (Power) | [katex]\text{watts (W)}[/katex] |
| [katex]p[/katex] (Momentum) | [katex]\text{kilogram meters per second (kgm/s)}[/katex] |
| [katex]\omega[/katex] (Angular Velocity) | [katex]\text{radians per second (rad/s)}[/katex] |
| [katex]\tau[/katex] (Torque) | [katex]\text{newton meters (Nm)}[/katex] |
| [katex]I[/katex] (Moment of Inertia) | [katex]\text{kilogram meter squared (kgm}^2\text{)}[/katex] |
| [katex]f[/katex] (Frequency) | [katex]\text{hertz (Hz)}[/katex] |
Metric Prefixes
Example of using unit analysis: Convert 5 kilometers to millimeters.
Start with the given measurement: [katex]\text{5 km}[/katex]
Use the conversion factors for kilometers to meters and meters to millimeters: [katex]\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}[/katex]
Perform the multiplication: [katex]\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}} = 5 \times 10^3 \times 10^3 \, \text{mm}[/katex]
Simplify to get the final answer: [katex]\boxed{5 \times 10^6 \, \text{mm}}[/katex]
Prefix | Symbol | Power of Ten | Equivalent |
|---|---|---|---|
Pico- | p | [katex]10^{-12}[/katex] | |
Nano- | n | [katex]10^{-9}[/katex] | |
Micro- | µ | [katex]10^{-6}[/katex] | |
Milli- | m | [katex]10^{-3}[/katex] | |
Centi- | c | [katex]10^{-2}[/katex] | |
Deci- | d | [katex]10^{-1}[/katex] | |
(Base unit) | – | [katex]10^{0}[/katex] | |
Deca- or Deka- | da | [katex]10^{1}[/katex] | |
Hecto- | h | [katex]10^{2}[/katex] | |
Kilo- | k | [katex]10^{3}[/katex] | |
Mega- | M | [katex]10^{6}[/katex] | |
Giga- | G | [katex]10^{9}[/katex] | |
Tera- | T | [katex]10^{12}[/katex] |
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